UNIVERSITY OF SOUTHAMPTON

SCHOOL OF PHYSICS AND ASTRONOMY

PHYS6003 Advanced Quantum Physics (AY 2022-23)
Problem Sheet 1
Note: solutions of the problems imply use of expressions given in the
lectures even when not explicitly recalled.
PROBLEM 1
Consider a vector space V and a generic vector | v i ∈ V . Show that the
axioms defining a vector space imply the useful relations:
• 0 |ui = 0 ;
• | − v i = −| v i .
Solution. Starting from the definition of inverse vector we can write
|vi + | − vi = 0. (1)
(i) We can then write in the left-hand side | v i+| −v i = (0+1) | v i+| −v i =
0| v i, obtaining 0| v i = 0.
(ii) We can now go back to Eq. (1) and recast the null vector on the right-
hand side as 0 = (1 − 1)| v i and using distributivity (1 − 1)| v i = | v i − | v i,
so that we obtain
|vi + | − vi = |vi − |vi. (2)
Add now on both sides +| − v i and use again on both sides | v i + | − v i = 0.
Considering that the null vector does not change a vector when added to it,
then it can be removed on both sides and in this way we finally proved our
thesis.
PROBLEM 2
Consider the three dimensional complex vector space of column vectors.
Determine whether or not the following column vectors form a linearly inde-
pendent set. You have to show whether imposing
a1 | u1 i + a2 | u2 i + a3 | u3 i = 0
implies necessarily a1 = a2 = a3 = 0 or not:
     
i 1 0
|u1 i =  0  , |u2 i =  1  , |u3 i =  2  .
−i 0 0
NOTE: You can use alternative methods as a cross-check, but you need
first to use the definition of linear dependence and linear independence as
requested by the problem (and discussed in the lectures).
Solution. Let us impose
       
i 1 0 0
a1  0  + a2  1  + a3  2  =  0  . (3)
−i 0 0 0

We find the simple system of three linear equations

i a1 + a2 = 0 (4)
a2 + 2 a3 = 0 (5)
−i a1 = 0 , (6)

with the trivial solution a1 = a2 = a3 = 0. Therefore, the three vectors are

linearly independent.

PROBLEM 3

(i) Calculate all inner products one can take with the three column vectors
| u1 i, | u2 i and | u3 i considered in the previous problem.
Solution. One finds:

h u1 | u1 i = ||u1 ||2 = 2 , h u1 | u2 i = −i , h u1 | u3 i = 0 ,

h u2 | u2 i = ||u2 ||2 = 2 , h u2 | u1 i = +i , h u2 | u3 i = 2 ,
h u3 | u3 i = ||u3 ||2 = 4 , h u3 | u1 i = 0 , h u3 | u2 i = 2 .
Notice that in all cases one has correctly h ui | uj i = h uj | ui i? .

(ii) Do they form an orthonormal basis?

Solution. They don’t, since the inner products do not satisfy h ui | uj i =
δij ;

(iii) Do they form a basis?

Solution. Yes, they do, since a set of n linearly independent vectors in
a n-dimensional vectors form a basis, in our case of course n = 3.
 UNIVERSITY OF SOUTHAMPTON
SCHOOL OF PHYSICS AND ASTRONOMY
PHYS6003 Advanced Quantum Physics (AY 2022-23)
Problem Sheet 2

Note: solutions of the problems imply use of expressions given in the

lectures even when not explicitly recalled.
PROBLEM 1
Consider the Gram-Schmidt procedure in a n-dimensional vector space
and discussed in class and lecture notes. What happens if you go one step
further and try to built a n + 1 basis vector | n + 1 i that is orthogonal to
all the mutually orthogonal unit norm vectors | i i for i = 1, . . . , n? Motivate
your answer.
Solution. Let us add to our set of vectors | e1 i, | e2 i, . . . , | en i an n + 1-
th vector | en+1 i. Since the space is n-dimensional we know that this vector
cannot be linearly independent of the other n vectors. If we now try to ignore
this and build a vector | (n + 1)0 i that is orthogonal to all the other mutually
orthogonal vectors | i i we built, then we should write
n
X
0
| (n + 1) i = | en+1 i − | i ih i | en+1 i .
i=1

However, since the n vectors | i i we built necessarily provide a basis and

since P
the expansion of a vector in a basis is unique, then necessarily one has
that ni=1 | i ih i | en+1 i = | en+1 i and, therefore, necessarily | (n + 1)0 i = 0.
As expected one does not manage to build a n + 1-th orthogonal vector of
unit norm.
PROBLEM 2
Prove the Triangle Inequality stating that the length of the sum of two
vectors (e.g., | u i and | v i) cannot exceed the sum of the lengths, explicitly

||u + v|| ≤ ||u|| + ||v|| .

Solution. Let us start from the definition of norm (squared) of a vector

applied to | u + v i:
||u + v||2 = h u + v | u + v i . (7)
Using then the properties of the linearity of the inner product we can write

||u + v||2 = ||u||2 + ||v||2 + h u | v i + h v | u i . (8)

Since h v | u i = h u | v i? and since for a complex number z one has z + z ? =

2Rez, we can then write

||u + v||2 = ||u||2 + ||v||2 + 2Reh u | v i . (9)

We can then use the simple inequality Rez ≤ |z| obtaining

||u + v||2 ≤ ||u||2 + ||v||2 + 2 |h u | v i| . (10)

Finally, using the Cauchy-Schwarz inequality we can write |h u | v i| ≤ ||u|| ||v||

and from this one finds

||u + v||2 ≤ ||u||2 + ||v||2 + 2 ||u|| ||v|| = (||u|| + ||v||)2 , (11)

and, finally, taking the square root one finds the Triangle Inequality.
PROBLEM 3
Prove the following useful identities:
h i h i h i
Ô, P̂ Q̂ = P̂ Ô, Q̂ + Ô, P̂ Q̂ (12)
h i h i h i
Ô P̂ , Q̂ = Ô P̂ , Q̂ + Ô, Q̂ P̂ . (13)

Solution. Let us start from the first identity. This can be derived with
straightforward steps using the definition of commutator and adding and
subtracting the term P̂ ÔQ̂:
h i
Ô, P̂ Q̂ = ÔP̂ Q̂ − P̂ Q̂Ô (14)
= P̂ ÔQ̂ − P̂ ÔQ̂ + ÔP̂ Q̂ − P̂ Q̂Ô
h i h i
= P̂ Ô, Q̂ + Ô, P̂ Q̂ .

The second indentity can be proved in an analogous way:

h i
Ô P̂ , Q̂ = ÔP̂ Q̂ − Q̂ÔP̂ (15)
= ÔP̂ Q̂ − ÔQ̂P̂ + ÔQ̂P̂ − Q̂ÔP̂
h i h i
= Ô P̂ , Q̂ + Ô, Q̂ P̂ .
 PROBLEM 4
Find the adjoint of the following linear combination of kets

α|ui + β ∗ |vi + ÔP̂ |wi + Ô |1ih1|Û |2i , (16)

where α and β ∗ are scalars and Ô, P̂ and Û are operators.

Solution. To obtain the adjoint, perform the operation of complex conju-
gation on the scalars, swap bras with kets, Ô → Ô† , P̂ → P̂ † , Û → Û † , and
reverse the order of the terms. The adjoint is then:

hu|α∗ + hv|β + hw|P̂ † Ô† + h2|Û † |1ih1|Ô† .

 UNIVERSITY OF SOUTHAMPTON
SCHOOL OF PHYSICS AND ASTRONOMY
PHYS6003 Advanced Quantum Physics (AY 2022-23)
Problem Sheet 3

Note: solutions of the problems imply use of expressions given in the

lectures even when not explicitly recalled.
PROBLEM 1
Suppose that the set of vectors {|1i,|2i,|3i} form an orthonormal basis for
a vector space, and α and β are scalars. For each of the following operators,
show whether it is a Hermitian operator or not:
a) |1ih2| + i|2ih1|
b) (α|1i + β ∗ |2i)† (β|1i − α∗ |2i)|3ih2| + |3ih3|
Solution. One needs to calculate the adjoint and check whether it remains
equal to the operator itself.
a) Let us define Ô = |1ih2| + i|2ih1|. In this case the adjoint is given by
†
Ô = |2ih1| − i|1ih2| = −i Ô and, therefore, the operator Ô is not Hermitian.
b) Let us define Ô = Ô1 + Ô2 with Ô1 = (α|1i+β ∗ |2i)† (β|1i−α∗ |2i)|3ih2|
and Ô2 = |3ih3|. The second operator Ô2 is clearly Hermitian since Ô2† =
(|3ih3|)† = |3ih3| = Ô2 . The first operator is tricky, and one had to work it
out to discover that, thanks to the orthonormality of the vector basis, it is
actually vanishing since one has:

Ô1 = (α|1i + β ∗ |2i)† (β|1i − α∗ |2i)|3ih2| (17)

= (h 1 |α? + h 2 |β) (β|1i − α∗ |2i)|3ih2| (18)
= (a? β − β α? ) |3ih2| (19)
= 0. (20)

Therefore, one has Ô† = Ô2† = Ô2 = Ô showing that Ô is Hermitian.

PROBLEM 2
Consider the 6 permutation operators P̂ijk (i 6= j 6= k and i, j, k = 1, 2, 3)
in a 3-dimensional vector space. Find their representing matrices
Solution. By the general definition given in the notes and in class and
specialised to the case n = 3 one has:

P̂ijk | 1 i = | i i P̂ijk | 2 i = | j i , P̂ijk | 3 i = | k i . (21)

The simplest case is clearly given by the case (i, j, k) = (1, 2, 3) since of course
in this case P̂123 = Iˆ and therefore it is represented by the identity matrix.
Let us then consider first the cyclic permutation P̂312 . In this case one has,
by definition of permutation operator,

P̂312 | 1 i = | 3 i , P̂312 | 2 i = | 1 i , P̂312 | 3 i = | 2 i .

In this way one finds for the matrix elements of the first row

h 1 |P̂312 | 1 i = 0 , h 1 |P̂312 | 2 i = 1 , h 1 |P̂312 | 3 i = 0 ,

of the second row

h 2 |P̂312 | 1 i = 0 , h 2 |P̂312 | 2 i = 0 , h 2 |P̂312 | 3 i = 1 ,

and of the third row

h 3 |P̂312 | 1 i = 1 , h 3 |P̂312 | 2 i = 0 , h 3 |P̂312 | 3 i = 0 ,

so that finally we can write

 
0 1 0
P312 = 0 0 1 .
1 0 0
One can of course repeat the same procedure to find the other four matrices.
However, notice that in this case what we obtained is that the permuta-
tion matrix is obtained from the identity matrix just simply permuting the
columns accordingly, i.e., the first column of the identity moves to the second
column, the second column to the third and the third to the first. Therefore,
following the same prescription one finds for the other four matrices:
   
0 0 1 0 1 0
P231 =  1 0 0  , P213 =  1 0 0  ,
0 1 0 0 0 1
   
1 0 0 0 0 1
P132 =  0 0 1  , P321 =  0 1 0  .
0 1 0 1 0 0
Notice that a compact summarising expression for the matrix elements for
all six permutation matrices is given by:

h ` |P̂ijk | m i = δ`i δm1 + δ`j δm2 + δ`k δm3

and this is equivalent to write in matrix form:
 
δ1i δ1j δ1k
P̂ijk → Pijk =  δ2i δ2j δ2k  .
δ3i δ3j δ3k

PROBLEM 3
We proved in the lecture that there always exists an operator relating two
orthonormal bases |i0 i and |ii in a way that:
|i0 i = Û |ii for all i = 1, . . . , n .
It was then shown that Û has to be a unitary operator.
i) Showing your working, find the explicit form of this unitary operator Û in
outer product notation.
ii) By using your answer in i) and the completeness relation, show that Û is
indeed unitary.
Note: You need to use your answer in part i), not the method discussed in
the lectures and on the notes.
Solution.
i) The operator is Û = ni=1 |i0 ihi|, and this can be derived very straight-
P
forwardly using the completeness relation:
X n n
X
ˆ
Û = Û I = Û | i ih i | = | i0 ih i | .
i=1 i=1

This result is confirmed by acting this operator on a particular ket |ji

(for fixed j) as follows:
Xn n
X
0
|i ihi|ji = |i0 iδij = |j 0 i .
i=1 i=1

Hence Û |ji = |j 0 i for a particular value of j, and hence Û |ii = |i0 i for
all i = 1 to n.
ii) One needs to show that Û Û † = Iˆ (all unitary operators must satisfy
this property by Pndefinition). Using the rules for forming the adjoint
† 0
one has Û = i=1 |jihj | with the summation index being j. Hence
Û Û † is given by:
Xn Xn Xn X n n
X
†
Û Û = 0 0
(|i ihi|) (|jihj |) = 0 0
|i ihj |δij = |i0 ihi0 | = Iˆ .
i=1 j=1 i=1 j=1 i=1
 UNIVERSITY OF SOUTHAMPTON
SCHOOL OF PHYSICS AND ASTRONOMY
PHYS6003 Advanced Quantum Physics (AY 2022-23)
Problem Sheet 4
(assigned Thursday 27/10 at 9am; deadline Friday 4/11 at 2pm)

Note: solutions of the problems imply use of expressions given in the

lectures even when not explicitly recalled.

PROBLEM 1
Consider the Euclidean space with an orthonormal basis | 1 i, | 2 i, | 3 i,
corresponding respectively to unit norm vectors along the i, j and k axes.
Consider the +π/2 rotation operator around the i axis, denoted by R̂i (π/2).
First, properly extending the Euclidean space, find all representations of the
eigenkets | ω = −i i, i.e., the eigenkets associated to the eigenvalue ω = −i,
of the operator R̂i (π/2). Second, specify those representations normalised to
unity.

[3]

Solution. One needs first of all to find the matrix representing the op-
erator R̂i (π/2) in the given basis. This is quite simple considering how the
operator acts on the three basis vectors:

R̂i (π/2)| 1 i = | 1 i (22)

R̂i (π/2)| 2 i = | 3 i (23)
R̂i (π/2)| 3 i = −| 2 i . (24)

Calculating all matrix elements h i |R̂i (π/2)| j i, one easily obtains the matrix
representation:  
1 0 0
R̂i (π/2) →  0 0 −1  .
0 1 0
One can immediately verify that, imposing det[R̂i (π/2) − ω I] = 0, implying
the characteristic equation (1 − ω) (ω 2 + 1) = 0, the three roots are ω1 = +1,
ω2 = +i, ω3 = −i.
Notice that if one strictly deals with the real Euclidean space, then there is
only one real eigenvalue and only one real eigenvector belonging to the space.
Therefore, in order to deal with complex eigenvalues, we necessarily have
to extend the Euclidean space to the case when scalars, and therefore also
eigenvalues, can be complex. Let us represent | ω3 i denoting its components
by u1 , u2 , u3 so that:  
u1
| − i i →  u2  .
u3
Let us then calculate | ω3 i as requested. One needs to impose
    
1 0 0 u1 u1
 0 0 −1   u2  = −i  u2  ,
0 1 0 u3 u3

implying

u1 = −iu1 ⇒ u1 = 0
−u3 = −i u2
.u2 = −i u3 .

Choosing u2 = α, where α is some arbitrary complex number, one has u3 =

+i α so that all representations can be written in the form
 
0
| − ii → α  1 .
i
√
Normalising the column vectors to unity requires α = eiϕ / 2, where ϕ is an
arbitrary phase, obtaining now:
 
iϕ 0
e  
| − ii → √ 1 .
2 i

Notice that there are still an infinite number of normalised column vectors
representing | − i i. This is because the overall phase ϕ is arbitrary. Of
course a simple choice could be simply ϕ = 0.
 PROBLEM 2
Is the outer product of a generic bra hu| with a generic ket |vi an Hermi-
tian operator? Motivate your answer.
[2]
Solution. No, in general it is not. The transformed ket under the action
of Ô ≡ | v ih u | is given by | ψ 0 i = | v ih u | ψ i. If we now consider the adjoint
bra we obtain:
h ψ 0 | = h ψ | u ih v | , (25)
showing that Ô† = | u ih v | and this implies that, in general, Ô† 6= Ô. Of
course in the special case | v i = | u i, where | u i is the adjoint of h u |, then
one has Ô† = Ô, i.e., h u | u i is Hermitian. We already know this, since
| u ih u | = h u | u i P̂u , i.e., it is the projector operator onto | u i times the
squared norm of | u i.

PROBLEM 3
Prove that the trace of an operator Ô, denoted by Tr(Ô), is an invariant
quantity, i.e., basis independent.
Solution 1.
This is the most straightforward way to show it, since it does not require
to indicate explicitly the operator Û transforming the vectors from one basis
to another. One simply has using the completeness relation first in the old
basis and then inversely in the new basis:
n
X
Tr(Ô)0 = h i0 |Ô| i0 i (26)
i=1
1,n
X
= h i0 | j ih j |Ô| i0 i
i,j
1,n
X
= h j |Ô| i0 ih i0 | j i
i,j
n n
!
X X
= h j |Ô | i0 ih i0 |j i
i=1 i=1
Xn
= h j |Ô| j i
i=1

= Tr(Ô) .
 Solution 2.
As written in the notes, one can also proceed first showing

X
Tr(Â B̂) = h i |Â B̂| i i (27)
i
X
= h i |Â| j i h j |B̂| i i (28)
i,j
X
= h j |B̂| i i h i |Â| j i (29)
i,j
X
= h j |B̂ Â| j i (30)
j

= Tr(B̂ Â) , (31)

where we used the completeness relation first directly and then inversely. We
can then use the unitary transformation Û to write:

n
X
0
Tr(Ô) = h i0 |Ô| i0 i (32)
i=1
Xn
= h i |Û † ÔÛ | i i
i=1
n
X
= h i |Û † Û Ô| i i
i=1
n
X
= h i |Ô| i i
i=1

= Tr(Ô) .

Solution 3. This is a kind of ‘brute force’ solution where the two steps
of Solution 2 are merged together and to this extent one has to use the
completeness relation twice in the same expression:
n
X
0
Tr(Ô) = h i0 |Ô| i0 i (33)
i=1
Xn
= h i |Û † ÔÛ | i i
i=1
X
= h i |Û † | j ih j |Ô| k ih k |Û | i i
i,j,k
X X
= h j |Ô| k i h k |Û | i i h i |Û † | j i
j,k i
X
= h j |Ô| k i δkj
j,k
X
= h j |Ô| j i = Tr(Ô) .
j

The operator Û is of course the unitary operator that realises the basis trans-
formation. Notice that in the third step we used the completeness relation
twice. We used the completeness relation in the fifth step as well, but in
reversed order on the index i.
PROBLEM 4
Consider the Hermitian operator Ô represented by the Hermitian1 matrix
 
0 0 1
O= 0 0 0 .
1 0 0

Find the unitary matrix U , representing the operator Û , that diagonalises

O, i.e., such that U † O U is a diagonal matrix.
[3]
Solution. We have seen that if a matrix O is Hermitian, then there exists
always a unitary matrix U diagonalising the Hermitian matrix, i.e., such
that (U † OU )ij = ωi δij , where the ωi ’s are the eigenvalues of O. Moreover
the columns of the unitary matrix are given by the (normalised) column
eigenvectors representing the | ωi i’s.
1
Notice that since it is real, it is also symmetric.
 Let us then solve the eigenvalue problem for the matrix O. From the
characteristic equation det(O − ω I) = 0 we find explicitly −ω 3 + ω = 0 so
that the eigenvalues are given by ω1 = −1, ω2 = 0 and ω3 = +1. Imposing
then Ô | ωi i = ωi ω (ωi ) , where we are denoting by ω (ωi ) the column vector
representing the eigenket | ωi i, one finds
     
1 0 1
(−1) 1  (0) (+1) 1  
| −1 i → ω =√ 0  , |0i → ω =  1  , | +1 i → ω =√ 0 .
2 −1 0 2 1
(34)
so that finally one arrives to the unitary matrix
 √ √ 
1/ 2 0 1/ 2
U = 0√ 1 0√  . (35)
−1/ 2 0 1/ 2

One can then straightforwardly verify that this does indeed diagonalise O:
 √ √   √ √ 
1/ 2 0 −1/ 2 0 0 1 1/ 2 0 1/ 2
U † OU =  0√ 1 0√   0 0 0   0√ 1 0√ 
1/ 2 0 1/ 2 1 0 0 −1/ 2 0 1/ 2
 
−1 0 0
=  0 0 0 .
0 0 +1
 UNIVERSITY OF SOUTHAMPTON
SCHOOL OF PHYSICS AND ASTRONOMY
PHYS6003 Advanced Quantum Physics (AY 2022-23)
Problem Sheet 5

Note: solutions of the problems imply use of expressions given in the

lectures even when not explicitly recalled.
PROBLEM 1
Consider the operator Ô represented by the Hermitian matrix2
 
0 0 i
 0 0 0  (36)
−i 0 0

Find the unitary matrix U , representing the operator Û , that diagonalises

O, i.e., such that U † O U is a diagonal matrix.
Solution. We have to repeat exactly the same procedure as in the problem
4 in PS n. 4. The characteristic equation and the eigenvalues are actually
the same, but not the normalised eigenvectors that are now given by:
     
1 0 1
1 1
| − 1i → √  0  , |0i →  1  , | + 1i → √  0  ,
2 +i 0 2 −i
so that one obtains the unitary matrix
 √ √ 
1/ 2 0 1/ 2
U = 0√ 1 0√  . (37)
+i/ 2 0 −i/ 2
One can then straightforwardly verify that this matrix does indeed diago-
nalise O:
 √ √   √ √ 
1/ 2 0 −i/ 2 0 0 1 1/ 2 0 1/ 2
U † O U =  0√ 1 0√   0 0 0  0√ 1 0√ 
1/ 2 0 +i/ 2 1 0 0 +i/ 2 0 −i/ 2
 
−1 0 0
=  0 0 0 .
0 0 +1
2
Notice that it is not real and, therefore, it is not symmetric.
 PROBLEM 2
Consider two operators Ô and P̂ , represented respectively by the matrices
   
1 0 1 2 1 1
O= 0 0 0  and P =  1 0 −1  . (38)
1 0 1 1 −1 2

Are they simultaneously diagonalisable? Motivate your answer showing the

derivation of the result.
Solution. The two matrices are Hermitian (they are also symmetric since
they are real). If we calculate the commutator of the two matrices, we
discover that it vanishes:
       
1 0 1 2 1 1 3 0 3 3 0 3
 0 0 0  ,  1 0 −1  =  0 0 0  −  0 0 0  = 0 .
1 0 1 1 −1 2 3 0 3 3 0 3
(39)
Since, as we have shown, if two Hermitian matrices commute with each oth-
ers, then there exists a common eigenbasis and a common unitary transfor-
mation that diagonalises them simultaneously, the answer is positive: they
are simultaneously diagonalisable.
PROBLEM 3
Derive the relation h i
x̂, k̂ = i , (40)

for the commutator of x̂ and k̂ (operators conjugate of each other) given in

the lecture notes.
Suggestion: You should start from
h i
h x | x̂, k̂ | ψ i , (41)

where | ψ i is a generic state vector.

Solution. Let us work in the x̂-basis. In this case one has h x |k̂| ψ i =
−idψ/dx. Let us now calculate
h i
h x | x̂, k̂ | ψ i ,

where | ψ i is a generic state vector and k̂ = −i D̂, where D̂ is the differ-

ential operator. Since h x |x̂ = x h x |, x̂| ψ i = | x ψ i and, by definition of
differential operator D̂, one has h x |D̂| ψ i = dψ(x)/dx, we obtain
h i
h x | x̂, k̂ | ψ i = −i x h x |D̂| ψ i + i h x |D̂| x ψ i , (42)
dψ d(x ψ)
= −i x +i
dx dx
dψ dψ
= −i x + ix + ihx|ψi
dx dx
= ihx|ψi.

Finally, since | ψ i is generic, one can indeed conclude that [x̂, k̂] = i.
 UNIVERSITY OF SOUTHAMPTON
SCHOOL OF PHYSICS AND ASTRONOMY
PHYS6003 Advanced Quantum Physics (AY 2022-23)
Problem Sheet 6

Note: solutions of the problems imply use of expressions given in the lectures
even when not explicitly recalled.
PROBLEM 1
Consider an observable represented by a Hermitian operator Ω̂ acting on
state vectors in a two dimensional Hilbert space. The eigenkets |ω1 i and |ω2 i
have distinct eigenvalues ω1 and ω2 respectively and form an orthonormal
basis. Consider a quantum ensemble of an unknown normalised state |ψi. A
measurement of the observable associated with Ω̂ is performed on each copy
| ψ i of the quantum ensemble, and it is found that 25% of the outcomes of
the measurements are the eigenvalue ω1 .
i) Derive an expression for the state |ψi (i.e. the state vector before the
measurement of Ω̂) in terms of the eigenvectors |ω1 i and |ω2 i.

Solution. The eigenvectors | ω1 i and | ω2 i provide an orthonormal ba-

sis for the two dimensional Hilbert space. We can, therefore, write for the
unknown normalised state | ψ i:

| ψ i = α | ω1 i + β | ω2 i ,

with |α|2 + |β|2 = 1. Since the probability to find ω1 as outcome is 0.25, then
we have |α|2 = 1/4 and |β|2 = 3/4 and including unknown phase factors we
can write √
1 i ϕ1 3 i ϕ2
| ψ i = e | ω1 i + e | ω2 i .
2 2
Notice that since physical observables are of the form |h λ | ψ i|2 , where | λ i
is some generic state like for example the eigenstate of another observable
that is not commuting with Ω̂, then only the phase difference φ = ϕ2 − ϕ1
can give physical effects and, without loss of generality, the state can also be
recast as √
1 3 iφ
| ψ i = | ω1 i + e | ω2 i .
2 2
ii) Give the expression for the expectation value hΩ̂i.
 Solution. This is simply given by
1 3
hΩ̂i = ω1 + ω2 .
4 4
PROBLEM 2
Consider the operators L̂x , L̂y and L̂z on a three-dimensional complex
Hilbert space, represented respectively by the following matrices:
 
0 1 0
1
Lx = √  1 0 1  , (43)
2 0 1 0
 
0 −i 0
1
Ly = √  i 0 −i  , (44)
2 0 i 0
 
1 0 0
Lz =  0 0 0  . (45)
0 0 −1

(i) What are the possible values one can obtain if L̂z is measured?
Solution. Since Lz is diagonal, then the eigenvalues are simply given
by the entries on the diagonal (-1, 0 and 1) and these give the possible
values that can be obtained if L̂z is measured.
(ii) Find the eigenvalues and the normalised eigenstates of L̂x in the L̂z -
basis;
Solution. The characteristic equation for the matrix Lx is simply
−ω(ω 2 − 1) = 0, so that the eigenvalues are Lx = −1, 0, +1. For
the normalised eigenstates one finds:
     
√1 1 1
√
1 1 1
| 1x i →  2  , | 0x i → √  0  , | −1x i →  − 2  .
2 2 2
1 −1 1
(46)
(iii) If the system is in the L̂z eigenstate with eigenvalue Lz = −1, and Lx
is measured, what are the possible outcomes and their probabilities?
Solution. One has to calculate the three probabilities |h Lx | − 1z i|2 ,
where | − 1z i → (0 0 1)T , finding P (Lx = 1) = 1/4, P (Lx = 0) = 1/2
and P (Lx = −1) = 1/4.
(iv) Consider the state | ψ i represented by the column vector
 
1/2
| ψ i →  1/2√
, (47)
1/ 2

in the L̂z -basis. If L̂2z is measured, in this state and a result +1 is

obtained, what is the state after the measurement? How probable was
this result? If L̂z is measured immediately afterwards, what are the
outcomes and respective probabilities? Motivate your answer showing
the derivation of the results.
Solution. L2z has eigenvalues +1 (with multiplicity two) and 0. Since L̂z
commutes with L̂2z , vectors in the subspace of the eigenvalue L2z = +1
can be expressed as linear combinations of the eigenvectors of L̂z with
Lz = ±1 that can be denoted by | 1z i and | − 1z i, or equivalently
one can say that | 1z i and | − 1z i provide a basis for the eigenspace
with L2z = +1. Notice that in this way L̂z breaks the degeneracy of
L2z = 1. The projector on this subspace is then given by P̂L2z =1 =
| 1z ih 1z | + | − 1z ih −1z | so that the projected state is given by
1 1
P̂L2z =1 | ψ i = | 1z i + √ | − 1z i , (48)
2 2
corresponding to the component of | ψ i on the plane orthogonal to the
eigenket with Lz = 0. When this projection is normalised, one obtains:
 √ 
1/ 3
0 P̂Lz =1 | ψ i
2
|ψ i = → 
p0
. (49)
||P̂L2z =1 ψ|| 2/3

The probability to obtain L2z = 1 is given by

P (L2z = 1) = h ψ |P̂L2z =1 | ψ i = |h 1z | ψ i|2 +|h −1z | ψ i|2 = 1/4+1/2 = 3/4 .

By measuring L̂z immediately after the measurement, one can only

obtain Lz = +1 with probability |h 1z | ψ 0 i|2 = 1/3 or Lz = −1 with
probability |h −1z | ψ 0 i|2 = 2/3. Notice that the measurement will nec-
essarily change the state. This is a nice example of two operators where
one (L̂2z ) has a degenerate eigenvalue and a second one (L̂z ) breaks its
degeneracy.
(v) The system is in a state | ψ i such that one has the following proba-
bilities to measure L̂z with the indicated values: P (Lz = −1) = 1/6,
P (Lz = 0) = 2/3, P (Lz = 1) = 1/6. Write down the most general, nor-
malised state | ψ i compatible with such probabilities in the L̂z -basis.
Solution. This is given by
r
1 2 1
| ψ i = eiα √ | 1z i + eiβ | 0z i + eiγ √ | − 1z i .
6 3 6
 UNIVERSITY OF SOUTHAMPTON
SCHOOL OF PHYSICS AND ASTRONOMY
PHYS6003 Advanced Quantum Physics (AY 2022-23)
Problem Sheet 7

Note: solutions of the problems imply use of expressions given in the lectures
even when not explicitly recalled.
PROBLEM 1.
Prove that the time-evolution operator Û (t, t0 ) has to be unitary, as stated
in the notes.
Solution. Let us start from the conservation of the total probability
X X
|h ωi | ψ(t0 i|2 = |h ωi | ψ(t i|2 .
i i

We can then rewrite LH side and RH side in the following way:

X X
h ψ(t0 ) | ωi ih ωi | ψ(t0 ) i = h ψ(t) | ωi ih ωi | ψ(t) i .
i i

Since both | ψ(t0 i and | ψ(t) i do not depend on the index i, we can bring
them outside the sum and at the same time, by definition of time-evolution
operator write | ψ(t) i = Û (t, t0 ) | ψ(t0 ) i, obtaining:
! !
X X
h ψ(t0 ) | | ωi ih ωi | | ψ(t0 ) i = h ψ(t0 ) |Û † (t, t0 ) | ωi ih ωi | Û (t, t0 )| ψ(t0 ) i .
i i

At this point one can use the completeness relation for the two sums, both
on the LH side and on the RH side, obtaining:

h ψ(t0 ) | ψ(t0 ) i = h ψ(t0 ) |Û † (t, t0 ) Û (t, t0 )| ψ(t0 ) i .

implying finally
Û † (t, t0 ) Û (t, t0 ) = Iˆ .
Notice that the the unitarity of Û (t, t0 ) implies of course that for a non-
normalised state vector | ψ(t) i the norm is conserved during the time evolu-
tion of the state, since a unitary transformation preserves inner products.
 PROBLEM 2.
Prove that the expression given in class for the time-evolution operator,

Ĥ
Û (t0 + dt, t0 ) = Iˆ − i dt
~
where dt is an infinitesimal time interval, does indeed satisfy the properties
of unitarity, continuity and composition if Ĥ is Hermitian.
Solution. It clearly satisfies continuity since

lim Û (t0 + dt, t0 ) = Iˆ .

dt→0

It also satisfies the composition property since

! !
Ĥ Ĥ
Û (t0 + dt1 + dt2 , t0 + dt1 ) Û (t0 + dt1 , t0 ) = Iˆ − i dt2 Iˆ − i dt1
~ ~
Ĥ
= Iˆ − i (dt1 + dt2 )
~
= Û (t0 + dt1 + dt2 , t0 ) .

Finally, we can see that

Ĥ † − Ĥ
Û † (t0 + dt, t0 ) Û (t0 + dt, t0 ) = Iˆ − i dt ,
~

showing that in order for Û (t0 + dt, t0 ) to satisfy unitarity, then necessarily
Ĥ † = Ĥ.
PROBLEM 3
Consider the case when the Hamiltonian operator does not depend ex-
plicitly on time.

• Find an expression for the time evolution operator as a function of the

Hamiltonian operator.
Solution. As we know a generic Hermitian operator Ω̂ can be written
in terms of its eigenvalues and its eigenkets | ωi i as (spectral decompo-
sition): X
Ω̂ = ωi | ωi i h ωi | . (50)
i
 The expression found for Û of course confirms that Û shares the same
eigenbasis as the Hamiltonian operator Ĥ and thus it commutes with
Ei
it. Even more its eigenvalues are functions (e−i ~ t ), of the eigenvalues
Ei of the Hamiltonian operator. Since we can always Taylor expand
a generic function f (Ei ) = i (Ein /n!) (df n /dEi )Ei =0 , and since the n-
P

th power of the Hamiltonian operator Ĥ n has eigenvalues Ein ,3 in this

case we can say that in general a generic operator Ô with eigenvalues
f (Ei ) can be expressed as Ô = f (Ĥ): in this way through the Taylor
series expansion one defines the function of an operator (of course this
requires that the series is convergent). In our specific case we have
E X  i t n E n
−i ~i t i
f (Ei ) = e = − ,
n
~ n!

(that is always convergent) so that we can write

Ĥ
Û = e−i ~
t
. (51)

Notice that this result could be obtained also from the Schrödinger
equation for Û (t, t0 ):
∂
i~ Û (t) = Ĥ Û (t) .
∂t
One could then be tempted to write straight away the solution of this
equation in the form of Eq. (51). However, Û is an operator, not just
a function of time and one cannot straight away apply results valid
for numbers, or in other words the solution (51) can be written just
symbolically, but one has to define rigorously what ‘exponential of Ĥ’
means. A way to proceed in a direct way is to consider Û (t0 + dt, t0 ):

Ĥ
Û (t0 + dt, t0 ) = Iˆ − i dt, ,
~
We can now obtain Û at a generic finite time t simply as the product of
the infinitesimal transformation Û (dt) for an infinite number N = t/dt
of times: !N
Ĥ t
Û (t) = lim 1 − i ,
N →∞ ~ N
3
This is easy to understand, for example: Ĥ 2 | Ei i = Ei2 | Ei i.
 and defining the exponential of Ĥ as
!N
Ĥ Ĥ t
e−i ~
t
= lim 1−i , (52)
N →∞ ~ N

one proves again the result.

• Write the expansion of the time evolution operator in the Hamiltonian

eigenbasis.
Solution. One has simply:
X Ei
Û (t) = e−i ~
t
| Ei i h Ei | ,
i

PROBLEM 4
Consider the three eigenkets |E1 i, |E2 i and |E3 i belonging to the or-
thonormal eigenbasis {Ei } of the Hamiltonian operator Ĥ, with i = 1, 2, . . . , ∞.
The Hamiltonian operator acts on the three eigenkets as follows:

Ĥ|E1 i = ~ω|E1 i; Ĥ|E2 i = 2~ω|E2 i; Ĥ|E3 i = 3~ω|E3 i.

A system at a particular time (t = 0) is in the following state (where A is a

real number):
r
7 i
|ψ(0)i = A | E1 i + | E2 i − √ | E3 i .
12 3

i) Write down the adjoint bra h ψ(0) |.

ii) Find A normalising | ψ(0) i to unity.

iii) Calculate the probability of obtaining an energy of 3~ω if a measure-

ment of energy is made on the state.

iv) If no measurement is made on | ψ(0) i, write down the expression for

| ψ(t) i in terms of {|E1 i, |E2 i, |E3 i} at a later time t > 0.

v) Using your answer in iv), explain whether the probability in iii) would
stay the same or be different at a later time t.
 Solution.

i) Considering that A is real, one has:

r
7 i
h ψ(0) | = h E1 | A + h E2 | + h E3 | √ .
12 3

to impose h ψ(0) | ψ(0) i = A2 + 7/12 + 4/12 = 1, obtaining

ii) One has √
A = ±1/ 12.

iii) This is simply given by P (E3 , t = 0) = |h E3 | ψ(0) i|2 = 1/3.

iv) The time evolution is described by phase factors containing the energy
eigenvalues multiplying each eigenket, in a way that (choosing positive
A):
r
1 7 i
| ψ(t) i = √ | E1 i e−iω t + | E2 i e−2 iω t − √ | E3 i e−3 iω t . (53)
12 12 3

v) It would stay the same since P (E3 , t) = |h E3 | ψ(t) i|2 = P (E3 , t =

0) = const. The reason is that the phase factor, where all the time
dependence resides, cancels out in this case.
 UNIVERSITY OF SOUTHAMPTON
SCHOOL OF PHYSICS AND ASTRONOMY
PHYS6003 Advanced Quantum Physics (AY 2022-23)
Problem Sheet 8
(assigned Wednesday 23 November 9am; deadline Friday 2 December 2pm)

Note: solutions of the problems imply use of expressions given in the lectures
even when not explicitly recalled.

PROBLEM 1
We have found, in the case of a two-dimensional space, that the time
evolution of the expectation value of a time-independent Hermitian operator
Q̂ for a generic state vector | ψ(t) i is given by:

α(t)
hQ̂i(t) = hQ̂i(0)−4 Re[a?1 (0) a2 (0) Q12 ] sin2 +2 Im[a?1 (0) a2 (0) Q12 ] sin α(t) .
2
(54)
Consider now the case of neutrino oscillations discussed on the notes and
denote the state vector by | ψ(t) i = | ν(t) i. At the initial time t = 0 this
coincides with the electron neutrino state: | ν(0) i = | νe i. Consider the
probability Pνe →νµ+τ (t) that the neutrino is detected on Earth as a tauon
or muon neutrino, described by the state vector | νµ+τ i, where t is the time
that takes to neutrinos to travel from the Sun to the Earth. This can be
expressed as Pνe →νµ+τ (t) = |h νµ+τ | ν(t) i|2 . The two flavour neutrino states
| νe i and | νµ+τ i are related to the mass eigenstates | ν1 i and | ν2 i by the
simple rotation

| νe i = cos θ12 | ν1 i − sin θ12 | ν2 i , (55)

| νµ+τ i = sin θ12 | ν1 i + cos θ12 | ν2 i ,

where θ12 is the solar neutrino mixing angle. The expression for Pνe →νµ+τ (t)
is found to be the famous two-neutrino oscillation probability formula:
 4 2

2 2 c ∆m12
Pνe →νµ+τ (t) = sin (2θ12 ) sin t , (56)
4cp~

where ∆m212 ≡ m22 − m12 . This can be derived from the expression (54)
writing Pνe →νµ+τ (t) in the form of a time dependent expectation value hQ̂i(t)
and properly specifying all the various involved quantities.
1. What is the operator Q̂ in this case?
Solution. The probability Pνe →νµ+τ (t) = |h νµ+τ | ν(t) i|2 can be recast
as
h ν(t) | νµ+τ ih νµ+τ | ν(t) i ,
so that one can make the identification Q̂ = | νµ+τ ih νµ+τ |.
2. What are the matrix elements Q11 , Q22 and Q12 ?
Solution. The matrix elements are defined as Qij ≡ h Ei |Q̂| Ej i where
in our case | E1 i = | ν1 i and | E2 i = | ν2 i. Therefore, one has simply
from Eqs. (55):
Q11 = h ν1 | νµ+τ ih νµ+τ | ν1 i = sin2 θ12 ,
Q12 = h ν1 | νµ+τ ih νµ+τ | ν2 i = sin θ12 cos θ12 ,
Q22 = h ν2 | νµ+τ ih νµ+τ | ν2 i = cos2 θ12 .

3. What are the two coefficients a1 (0) and a2 (0)?

Solution. Since ai (0) = h νi | ν(0) i, one has a1 (0) = h ν1 | νe i = cos θ12
and a2 (0) = h ν2 | ν(0) i = − sin θ12 .
4. Derive the expression (56) noticing that the imaginary part vanishes
for this particular application.
Solution. One can see straightforwardly that
hQ̂i(0) = h ν(0) | νµ+τ ih νµ+τ | ν(0) i = h νe | νµ+τ ih νµ+τ | νe i = 0 .

This is confirmed by calculating hQ̂i(0) using the general expression

(see notes)
hQ̂i(0) = |a1 (0)|2 Q11 + |a2 (0)|2 Q22 + 2Re [a?1 (0) a2 (0) Q12 ] ,
finding as expected
hQ̂i(0) = |a1 (0)|2 Q11 + |a2 (0)|2 Q22 + 2Re [a?1 (0) a2 (0) Q12 ]
= cos2 θ12 sin2 θ12 + cos2 θ12 sin2 θ12 − 2 cos2 θ12 sin2 θ12
= 0.
Therefore, considering also that all coefficients are real, the expression
(54) simplifies into:
α(t)
hQ̂i(t) = −4 a1 (0) a2 (0) Q12 sin2 ,
2
 where the time dependent phase is given by

E2 − E1 c4 ∆m212
α(t) = t= t.
~ 4cp~

Plugging all found expressions and using 4 sin2 θ12 cos2 θ12 = sin2 (2θ12 ),
one finally arrives to Eq. (56).

[4]

PROBLEM 2
Consider the quantum harmonic oscillator.

• The action of the raising operator on a generic eigenstate | n i is such

that
↠| n i = Cn+ | n + 1 i . (57)
√
Prove that Cn+ = n + 1, barring an arbitrary phase phase factor (i.e.,
choosing Cn+ to be real) and imposing h n | n i = 1 for all values of n.

[3]

Solution. Taking the squared norm of both left hand-side and right-
hand side in Eq. (57) and swapping them, one obtains:

|Cn+ |2 = h n |a† a | n i
= h n |[a† , a] + a† a| n i
= 1 + n,

where we used [a† , a] = 1 and the orthonormality of √

the | n i’s. In this
way, choosing Cn+ real and positive, one finds Cn+ = 1 + n.

• Analogously, we have seen that the lowering operator acts on an eigen-

state | n i, for n ≥ 1, in a way that

â | n i = Cn− | n − 1 i . (58)
√
Prove that Cn− = n, again barring an arbitrary phase factor and
imposing h n | n i = 1 for all values of n.
 Solution. It is analogous to the previous problem, even simpler since one
diractly obtains a† a without having to go through the commutator:

|Cn− |2 = h n |a† a | n i
= n,
√
In this way, choosing Cn− real, one finds Cn− = n.

[3]
 UNIVERSITY OF SOUTHAMPTON
SCHOOL OF PHYSICS AND ASTRONOMY
PHYS6003 Advanced Quantum Physics (AY 2022-23)
Problem Sheet 9

Note: solutions of the problems imply use of expressions given in the lectures
even when not explicitly recalled.
PROBLEM 1
Write the momentum operator p̂ in terms of â and ↠and evaluate the
expectation values hp̂i and hp̂2 i for a simple harmonic oscillator in the eigen-
state | n i. Finally, calculate the expectation value Kn ≡ h n |K̂| n i of the
kinetic energy operator K̂ = p̂2 /(2m) and compare it to the result obtained
for the expectation value of the potential energy Vn
Solution.
From the definitions of â and ↠, one easily finds
i
P̂ = − √ (â − ↠) (59)
2
and since P̂ is defined as
1
P̂ = √ p̂ ,
m~ω
one finds r
m~ω
p̂ = −i (â − ↠) .
2
From this expression we can now easily calculate the expectation value of p̂
on the eigenstates | n i:
hp̂i = h n |p̂| n i
r
m~ω
= i h n |(â − ↠)| n i
r 2
m~ω −
= i (Cn h n | n − 1 i − Cn+ h n | n + 1 i)
2
= 0,
where in the last step we used the orthonormality of the | n i’s .
Let us now calculate hp̂2 i. First of all from Eq. (59) we find
1
P̂ 2 = − (â â + ↠↠− ↠â − â ↠) (60)
2
and from this one immediately obtains
m~ω
hp̂2 i = h n |(↠â + â ↠)| n i
2  
1
= m~ω n + ,
2

where we used h n |â2 | n i = h n |(↠)2 | n i = 0. Notice that this result con-

firms, as expected, that the expectation value of the kinetic energy is equal to
the expectation value of the potential energy calculated in the lecture notes:

p̂2
 
~ω 1 1
Kn ≡ h n |K̂| n i = h n | |ni = n+ = En = Vn .
2m 2 2 2

PROBLEM 2
A particle is in a j = 1 eigenstate of the squared angular momentum
operator Jˆ2 . The matrix representation of Jˆy (i.e., the y-component of the
angular momentum vector operator) in the basis of common eigenvectors
{|1, mi} of Jˆ2 and Jˆz is given by:
 
0 −i 0
~
Jˆy = √  i 0 −i  .
2 0 i 0

(a) Show that the eigenvalues my ~ of Jˆy are given by +~, 0 and −~ corre-
sponding, respectively, to my = +1, 0 and −1.
Solution. The eigenvalues, denoted by λ, are found by imposing the
condition:
det(Jˆy − λI) = 0 ,
translating, in terms of matrix representations, into:
 
−λ −i~√
2
0
det  √i~2 −λ −i~ √  = 0 .
 
2
0 √i~2 −λ

This leads to the characteristic equation:

−λ3 + λ ~2 = 0 ⇔ λ(λ2 − ~2 ) = 0 ,
 showing that the eigenvalues are indeed λ = 0 and λ = ±~. Alterna-
tively, one can calculate directly the dimensionless quantum number
my . In this way ~ can be factorised out from the matrix and one finds
very easily my = 0, ±1 still yielding λ = 0, ±~.
(b) Find the column vector representations of the normalised eigenvectors
of Jˆy in the basis {| 1, m i}. When normalising, you may neglect overall
multiplicative phase factors which do not change the norm.
Solution.
Let us start from the eigenvector of Jˆy with eigenvalue +~:
    
0 −i 0 a a
~
√  i 0 −i   b  = ~  b  .
2 0 i 0 c c
This leads to the relations
√
b = i 2a, c = −a ,
and so we can write the eigenvector as
 
√a
| 1, 1y i →  i 2a  .
−a
To normalise we impose h 1, 1y | 1, 1y i = 1, finding:
 
√ √a
a∗ − i 2a∗ − a∗  i 2a  = 4|a|2 = 1 .

−a
Therefore, the normalised eigenvector (up to multiplication by an ar-
bitrary phase) can be finally written as:
 
√1
1
| 1, 1y i →  i 2  .
2
−1

Using the same method, one can obtain the normalised eigenvectors
for |1, 0y i and |1, −1y i (up to multiplication by an arbitrary phase),
finding:
   
1 1√
1 1
|1, 0y i → √  0  and |1, −1y i →  −i 2  .
2 2
1 −1
(c) If the particle is in the eigenstate described by | 1, 1y i, denoting the
normalised eigenvector of Jˆy with quantum number my = +1, evaluate
the probabilities of finding values +~, 0, −~ as outcomes of a measure-
ment of the Jˆx angular momentum operator.
NOTE: You will need to use the expressions for the normalised eigen-
vectors of Jˆx in the basis {|1, mi} given in the lecture notes (they can
be quoted without derivation).
Solution. Now let us consider the system in the state |1, 1y i and recall
that the normalised eigenvectors of Jˆx are:
     
√1 1 1
√
1 1 1
|1, 1x i →  2  , |1, 0x i = √  0  , |1, −1x i →  − 2  .
2 2 2
1 −1 1

The probability P (Jx = ~) can be calculated as |h1, 1x |1, 1y i|2 :

  2
√ √1
1
1 1
P (Jx = ~) = |h1, 1x |1, 1y i|2 = 1 2 1  i 2  = .
2 2 4
−1

A similar calculation for gives P (Jx = 0) = |h0x |1y i|2 = 0.5 and P (Jx =
−~) = |h−1x |1y i|2 = 0.25. Notice that the sum of all probabilities is 1,
as it MUST be (this is a sanity check).
 UNIVERSITY OF SOUTHAMPTON
SCHOOL OF PHYSICS AND ASTRONOMY
PHYS6003 Advanced Quantum Physics (AY 2022-23)
Problem Sheet 10

Note: solutions of the problems imply use of expressions given in the lectures
even when not explicitly recalled.

PROBLEM 1
Consider a spin 1/2 particle. By explicitly calculating the expectation
values of the spin operators Ŝx , Ŝy and Ŝz (denoted by hŜx i, hŜy i and hŜz i
respectively), show that it is impossible for the particle to be in a state of
spin such that hŜx i = hŜy i = hŜz i = 0.
Hint:
Write a generic (normalised) state of spin as
 
a
|ψs i = ,
b

where a and b are complex numbers and |a|2 + |b|2 = 1.

Solution. Let us recall that the matrices representing the spin components
can be conveniently expressed, in the Ŝz basis, in terms of the Pauli matrices
~σ = (σx , σy , σz ) as
~ = ~ ~σ ,
S (61)
2
where:
     
0 1 0 −i 1 0
σx = , σy = , σz = . (62)
1 0 i 0 0 −1

We can then conveniently work with the dimensionless Pauli matrices rep-
resenting, in the Ŝz basis, the dimensionless operators σ̂i = Ŝi /(~/2), thus
avoiding to work with the ~/2 factor. Let us start imposing the constraint
hŜz i = 0:
  
∗ ∗ 1 0 a
hψs |σ̂z |ψs i = (a b ) = |a|2 − |b|2 = 0 .
0 −1 b

We can always parameterise a = cos θ and b = sin θ eiφ , removing one overall
phase. The constraint |a|2 − |b|2 = 0 then gives cos 2θ = 0, that is satisfied
for θ = π/4. Let us now proceed imposing the constraint hσ̂y i = 0. One finds
  
∗ ∗ 0 −i a
hσ̂y i = hψs |σ̂y |ψs i = (a b )
i 0 b
? ?
= i (ab − ba )
= −2 Im(cos θ sin θ e−iφ )
= sin 2θ sin φ
= sin φ ,

so that the conditions becomes sin φ = 0 that immediately fixes φ = n π,

implying eiφ = ±1, so that we can write:
| + i ± |−i
| ψs i = √ . (63)
2
Finally, we can impose the constraint hσ̂x i = 0, finding:
  
1 0 1 1
hψs |σ̂x |ψs i = (1 ± 1) = ±1 .
2 1 0 ±1

This result is not suprising, since indeed (63) is the eigenstate of Ŝx with
eigenvalue ±~/2 (see equation on the notes). Hence, it is impossible for a
spin 1/2 particle to be in a state such that hŜx i = hŜy i = hŜz i = 0, one can
at most have two of these expectation values simultaneously vanishing but
not three. Notice that if we were first imposing hσ̂x i = 0 without having
imposed first hσ̂y i = 0, we would have found

hσ̂x i = cos φ = 0 , (64)

implying φ = π/2+nπ and in that case one would find, vice versa, hσ̂y i = ±1.
PROBLEM 2
Consider the spin component operator along a generic direction n ≡
(θ, φ), where θ is the polar angle and φ is the azimuthal angle:

Ŝn = nx Ŝx + ny Ŝy + nz Ŝz .

i) Solve the eigenvalue problem for Ŝn , calculating the eigenvalues and
the corresponding eigenvectors in terms of θ and φ.
ii) Specialise your general expressions for the eigenvectors to the three
cases when n̂ coincides with one of the three Cartesian axes.
 Solution.
i) Let us solve the eigenvalue problem for Ŝn :
1 1
Ŝn̂ | , mn i = mn ~| , mn i . (65)
2 2
~ˆ
ˆ ≡ S/(~/2)
It is again convenient to work with the dimensionless operator ~σ
and solve instead the eigenvalue problem for
ˆ · n = nx σ̂x + ny σy + nz σ̂z ,
σ̂n ≡ ~σ
so that Eq. (65) becomes
1 1
σ̂n | , mn i = 2 mn | , mn i . (66)
2 2
The operators σ̂i are represented by the Pauli matrices given in Eq. (62). In
this way one finds for the matrix representation of σ̂n :
 
nz nx − i ny
σn = . (67)
nx + i ny −nz
The components of the unit vector n are related to the polar angle θ and
azimuth angle φ by
(nx , ny , nz ) = (sin θ cos φ, sin θ sin φ, cos θ) , (68)
clearly respecting the normalisation condition n2x + n2y + n2z = 1. From the
characteristic equation
 
nz − 2 mn nx − i ny
det = −(n2z − (2 mn )2 ) − (n2x + n2y ) = 0 ,
nx + i ny −nz − 2 mn
one immediately finds mn = ±1/2, corresponding to Sn = ±~/2, confirming
that the eigenvalues of the spin components do not depend on the particular
direction, as one would expect from space isotropy.
Let us now calculate the two eigenvectors first defining
 
1 1 a±
| ±n i ≡ | , mn = ± i → ,
2 2 b±

with |a± |2 + |b± |2 = 1, and then imposing

  
nz − 2 mn nx − i ny a±
= 0.
nx + i ny −nz − 2 mn b±
From this one finds the following condition on a± and b± :
a± (nz ∓ 1) = −b± (nx − i ny ) .
Using the relation (68), one then finds
b± sin θ = −a± (cos θ ∓ 1) ei φ . (69)
Rewriting a± = |a± | eiφa and b± = |b± | ei φb and using the simple trigonomet-
ric relations
θ θ
cos θ + 1 = 2 cos2 and cos θ − 1 = −2 sin2 ,
2 2
one then arrives to the conditions:
sin(θ/2) i φ
b+ ei (φb −φa ) = a+ e ,
cos(θ/2)
and
cos(θ/2) i φ
b− ei (φb −φa ) = −a− e .
sin(θ/2)
The simplest choices are then |a+ | = cos(θ/2), |b+ | = sin(θ/2), |a− | =
sin(θ/2), |b− | = cos(θ/2) and φa = 0 in both cases4 , so that one finally
finds:
 
cos(θ/2)
| +n i → , (70)
sin(θ/2) e+iφ
 
sin(θ/2)
| −n i → . (71)
− cos(θ/2) e+iφ
ii) We can now easily specialise these expressions to the cases n = i, j, k:
• The eigenvectors of Ŝx are found taking (nx , ny , nz ) = (1, 0, 0), corre-
sponding to θ = π/2 and φ = 0, so that one obtains:
 
1 1
| +x i → √ , (72)
2 1
 
1 1
| −x i → √ . (73)
2 −1
These reproduce the results given in the lecture notes.
4
Notice that the normalisation condition |a± |2 + |b± |2 = 1 implies we can write a± =
cos α± and b± = sin α± , in the first case the choice is equivalent to say α+ = θ/2 and in
the second case α− = θ/2 − π/2.
 • The eigenvectors of Ŝy are found taking (nx , ny , nz ) = (0, 1, 0), corre-
sponding to θ = π/2 and φ = π/2 so that one obtains:
 
1 1
| +y i → √ , (74)
2 i
 
1 1
| −y i → √ . (75)
2 −i

These reproduce the results in the lecture notes.

• Finally, we can find the eigenvectors of Ŝz taking nx (nx , ny , nz ) =

(0, 0, 1), corresponding to θ = 0 and φ = 0 and finding correctly:
 
1
|+i → , (76)
0
 
0
|−i → . (77)
1

PROBLEM 3
We have seen that the matrix representations S = (Sx , Sy , Sz ) for the spin
components of a spin 1/2 particle can be conveniently expressed in terms of
the Pauli matrices ~σ = (σx , σy , σz ) as
~
S= ~σ . (78)
2
We have also seen that the rotation operator for a finite rotation of an angle
θ about an axis n can be written as
Ĵ·n
D̂(n, θ) = e−i ~
θ
. (79)

For a spin 1/2 particle one has simply Ĵ = Ŝ and one finds for its (2 × 2)
matrix representation (as usual, in the Ŝz -basis):
   
−i S·n θ θ θ
e ~ = cos I − i sin n · ~σ . (80)
2 2
• Consider the generic spinor represented by
 
χ+
χ= . (81)
χ−
 Derive the rotated spinor χ0 in the case of a rotation 2π around the
generic axis n. Comment on the obtained result.
Solution If we take θ = 2π then simply
Ĵ·n
D̂(n, 2π) = e−i ~
2π
= −I , (82)

so that
D̂(n,2π)
χ −→ χ0 = −χ . (83)
This shows that a spinor is not singled valued since it changes sign
under a 2π rotation. It is needs to be rotated by a 4π rotation to get
back to its original status.

• Consider now the spinor

 
1
|+i → . (84)
0

Derive the transformed spinors under a rotation −π/2 about the x̂-
axis and under a rotation +π/2 around the ŷ-axis. Comment on the
obtained results.
Solution A rotation of −π/2 about the x̂ axis corresponds to a rotation
operator in the s = 1/2 subspace represented by the matrix
 
1 1 i
D̂(i, −π/2) = √ , (85)
2 i 1
so that one finds
   
1 D̂(i,−π/2) 0 1 1
+= −→ + =√ = +y . (86)
0 2 i

Not surprisingly, we found that the spinor that represents the eigenstate
of Ŝy with Sy = +~/2: this is consistent with the fact that under such a
rotation a vector oriented along the z-axis is transformed into a vector
along the y-axis (with the same length).
Analogously, if we rotate the spinor + of an angle +π/2 about the
y-axis we have now
 
1 1 −1
D̂(j, +π/2) = √ , (87)
2 1 1
so that one finds
   
1 D̂(j,+π/2) 0 1 1
+= −→ + =√ = +x . (88)
0 2 1

Again, not surprisingly, we have found that a +π/2 rotation about the
y-axis transforms a | + i spinor into a spinor that is eigenstate of Ŝx ,
in the same way as a vector along the z-axis is rotated into a vector
along the x-axis.